The equation of a circle $C$ is $x^2+y^2-14x+6y+57 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-14x) + (y^2+6y) = -57$ $(x^2-14x+49) + (y^2+6y+9) = -57 + 49 + 9$ $(x-7)^{2} + (y+3)^{2} = 1 = 1^2$ Thus, $(h, k) = (7, -3)$ and $r = 1$.